EXERCISE 14.1

1. Complementary events – For any event \(E\) and its complement \(E'\) ("not E"), the two together cover the whole sample space. Hence, \[P(E) + P(E') = 1.\]
2. Impossible event – An event that cannot occur has probability zero. Such an event is called an *impossible event*.
\[P(\text{impossible event}) = 0.\]
3. Certain event – An event that is sure to occur has probability one. Such an event is called a *certain event*.
\[P(\text{certain event}) = 1.\]
4. Elementary events – The set of all elementary (or simple) events of an experiment is exhaustive; their probabilities add up to the total probability of the sample space, which is 1.
\[\sum_{i=1}^{n} P(E_i) = 1,\] where \(E_i\) are the elementary events.
5. Range of probability – By definition, probability of any event lies between 0 and 1 inclusive.
\[0 \le P(A) \le 1.\]
These statements directly answer the blanks in the question.

1. Identify the outcomes for each experiment and denote their probabilities.
- (i) Outcomes: "Car starts" (S) and "Car does not start" (N). Let \(P(S)=p\) and \(P(N)=1-p\).
- (ii) Outcomes: "Shot made" (M) and "Shot missed" (X). Let \(P(M)=q\) and \(P(X)=1-q\).
- (iii) Outcomes: "Answer right" (R) and "Answer wrong" (W). If the student guesses randomly, \(P(R)=\frac{1}{2}\) and \(P(W)=\frac{1}{2}\).
- (iv) Outcomes: "Boy" (B) and "Girl" (G). In the idealised model used in textbooks, \(P(B)=\frac{1}{2}\) and \(P(G)=\frac{1}{2}\).

2. Check whether the two outcomes have the same probability.
- (i) In real life the probability of a car starting is usually greater than the probability of it not starting ( \(p>\frac{1}{2}\) ). Hence the outcomes are not equally likely.
- (ii) For a basketball player the chance of making a shot depends on skill, distance, etc.; generally \(q
eq \frac{1}{2}\). Hence the outcomes are not equally likely.
- (iii) When a true‑false question is attempted by a student who does not know the answer and guesses, each of the two possibilities (right or wrong) has probability \(\frac{1}{2}\). Hence the outcomes are equally likely.
- (iv) Assuming the birth of a boy or a girl is equally probable (the textbook model), each outcome has probability \(\frac{1}{2}\). Hence the outcomes are equally likely.

3. Conclusion: The experiments (iii) and (iv) have equally likely outcomes, while (i) and (ii) do not.

4. Explanation: The definition of equally likely outcomes requires that the sample space consist of outcomes that occur with the same probability. In (i) and (ii) the underlying physical or skill‑related factors make one outcome more probable than the other, violating the condition. In (iii) (random guess) and (iv) (idealised birth of boy/girl) the two outcomes each have probability \(\frac{1}{2}\), satisfying the definition.

1. Identify the sample space: For a single toss of a fair coin, the set of elementary outcomes is \(S = \{H, T\}\).
2. Assign probabilities: Since the coin is unbiased, the probability of each outcome is
$$P(H) = P(T) = \frac{1}{2}.$$
3. Interpret the outcomes: Let Team A get the ball if the result is Head and Team B if the result is Tail (or vice‑versa).
4. Compute the probability for each team:
$$P(\text{Team A gets the ball}) = P(H) = \frac{1}{2},$$
$$P(\text{Team B gets the ball}) = P(T) = \frac{1}{2}.$$
5. Conclude fairness: Both teams have exactly the same probability (\(\frac12\)) of being selected. Since no team is favoured, the coin toss is a *fair* method of deciding which team gets the ball.
6. Link to definition of fairness: According to the definition of a fair experiment in statistics, an experiment is fair if all possible outcomes are equally likely. The coin toss satisfies this condition, therefore it is considered fair.

1. Write down the fundamental property of probability:
$$0 \leq P(E) \leq 1$$
2. Convert each option to a decimal (if needed) and compare with the interval.
- (A) $\frac{2}{3}=0.666\ldots$ → lies between 0 and 1 → valid.
- (B) $-1.5$ → less than 0 → invalid.
- (C) $15\% = 0.15$ → lies between 0 and 1 → valid.
- (D) $0.7$ → lies between 0 and 1 → valid.
3. The only option that violates the condition $0 \le P \le 1$ is (B).
4. Hence, (B) cannot be the probability of an event.

1. Write down the given probability:
$$P(E) = 0.05$$

2. Apply the complement rule:
$$P(\text{not }E) = 1 - P(E)$$

3. Substitute the given value:
$$P(\text{not }E) = 1 - 0.05$$

4. Perform the subtraction:
$$P(\text{not }E) = 0.95$$

Thus, the probability of the event ‘not E’ is \(0.95\).

Step 1: Identify the total number of possible outcomes. Since the bag contains only lemon flavoured candies, every candy drawn will be lemon. Hence, total outcomes = 1 (lemon).

Step 2: (i) Probability of drawing an orange flavoured candy.
- Favourable outcomes for orange = 0 (no orange candy in the bag).
- Using the definition, $$P(\text{orange}) = \frac{0}{1} = 0.$$ Therefore, the probability is 0.

Step 3: (ii) Probability of drawing a lemon flavoured candy.
- Favourable outcomes for lemon = 1 (the only candy present).
- $$P(\text{lemon}) = \frac{1}{1} = 1.$$ Hence, the probability is 1.

1. Assumption: There are 365 equally likely days in a year and birthdays are independent.

2. Probability that all three birthdays are different:
$$P(\text{all distinct}) = \frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}
= \frac{364\times 363}{365^{2}}\approx 0.992.$$
This matches the given value (0.992).

3. Complementary event: The event "at least two students have the same birthday" is the complement of "all three birthdays are different".
$$P(\text{at least two same}) = 1 - P(\text{all distinct}).$$

4. Compute the required probability:
$$P(\text{at least two same}) = 1 - 0.992 = 0.008.$$
Hence the probability that two students (among the three) share the same birthday is \(0.008\) (or \(8\times10^{-3}\)).

5. Answer in fraction (optional):
$$0.008 = \frac{8}{1000} = \frac{1}{125}.$$
So the required probability can also be expressed as \(\frac{1}{125}\).

Step 1: Identify the total number of balls in the bag.
$$\text{Total balls}=3\text{ (red)}+5\text{ (black)}=8.$$

Step 2: (i) Probability of drawing a red ball.
- Favourable outcomes = number of red balls = 3.
- Total outcomes = 8.
Thus,
$$P(\text{red})=\frac{\text{Number of red balls}}{\text{Total balls}}=\frac{3}{8}.$$

Step 3: (ii) Probability of drawing a ball that is not red (i.e., black).
- This is the complement of the event "red".
- Using the complement rule,
$$P(\text{not red})=1-P(\text{red})=1-\frac{3}{8}=\frac{5}{8}.$$

Alternatively, directly count the black balls:
Number of black balls = 5, so
$$P(\text{not red})=\frac{5}{8}.$$

Thus the required probabilities are $\frac{3}{8}$ and $\frac{5}{8}$ respectively.

1. Total number of marbles
$$N = 5\text{ (red)} + 8\text{ (white)} + 4\text{ (green)} = 17$$

2. (i) Probability of drawing a red marble
- Favourable outcomes = 5 (red marbles)
- $$P(\text{red}) = \frac{5}{17}$$

3. (ii) Probability of drawing a white marble
- Favourable outcomes = 8 (white marbles)
- $$P(\text{white}) = \frac{8}{17}$$

4. (iii) Probability of NOT drawing a green marble
- Marbles that are not green = red + white = 5 + 8 = 13
- $$P(\text{not green}) = \frac{13}{17}$$

All fractions are already in their simplest form.

1. Count the total number of coins\
\[\text{Total coins}=100+50+20+10=180\]\
2. (i) Probability of getting a 50p coin\
- Favourable outcomes = number of 50p coins = 100.\
- Hence, \[P(\text{50p coin}) = \frac{100}{180}=\frac{10}{18}=\frac{5}{9}.\]\
3. (ii) Probability that the coin is NOT a \`5 coin\
- Coins that are not \`5 = total coins – \`5 coins = 180 – 10 = 170.\
- Hence, \[P(\text{not \`5 coin}) = \frac{170}{180}=\frac{17}{18}.\]

1. Identify the total number of fish in the tank.
$$\text{Total fish}=5\text{ (male)}+8\text{ (female)}=13.$$

2. Identify the number of favourable outcomes (choosing a male fish).
$$\text{Favourable outcomes}=5\text{ (male fish)}.$$

3. Apply the definition of probability:
$$P(\text{male fish})=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{5}{13}.$$

4. Since the fish is taken out at random, each fish is equally likely to be chosen, so the probability computed above is the required answer.

1. Identify the sample space\
The bag contains 5 red balls and 6 blue balls, making a total of \(5+6 = 11\) balls. Each ball is distinct and equally likely to be drawn.

2. Probability of each individual ball\
Since the balls are equally likely, the probability of drawing any particular ball (say, the first red ball) is \[P(\text{specific ball}) = \frac{1}{11}.\]

3. Define the events of interest\
- Event \(R\): "A red ball is drawn" (favourable outcomes = 5).\
- Event \(B\): "A blue ball is drawn" (favourable outcomes = 6).

4. Compute the probabilities of the events\
\[P(R) = \frac{\text{number of red balls}}{\text{total balls}} = \frac{5}{11}.\]
\[P(B) = \frac{\text{number of blue balls}}{\text{total balls}} = \frac{6}{11}.\]

5. Analyse the given argument\
The statement "each of them has a probability \(\frac{1}{11}\)" is correct only when "them" refers to each *individual ball*. It is incorrect if "them" is meant to denote the two *events* (red or blue). The events have different numbers of favourable outcomes, so their probabilities are \(\frac{5}{11}\) and \(\frac{6}{11}\) respectively, not \(\frac{1}{11}\).

6. Conclusion\
The argument is partially right (for individual balls) but wrong for the events of drawing a red or a blue ball. Hence we do not agree with the claim that each colour has probability \(\frac{1}{11}\).

1. Identify the sample space\\
The arrow can stop at any of the eight numbers: $$S=\{1,2,3,4,5,6,7,8\}.$$\\
Hence, total number of equally likely outcomes \(n(S) = 8\).

2. (i) Probability of pointing at 8\\
- Favourable outcome: \{8\} → \(n = 1\).
- $$P(8)=\frac{1}{8}.$$\\
3. (ii) Probability of pointing at an odd number\\
- Odd numbers in \(S\): \{1,3,5,7\} → \(n = 4\).
- $$P(\text{odd})=\frac{4}{8}=\frac{1}{2}.$$\\
4. (iii) Probability of pointing at a number greater than 2\\
- Numbers greater than 2: \{3,4,5,6,7,8\} → \(n = 6\).
- $$P(>2)=\frac{6}{8}=\frac{3}{4}.$$\\
5. (iv) Probability of pointing at a number less than 9\\
- All numbers 1 to 8 satisfy this condition, so favourable outcomes = 8.
- $$P(<9)=\frac{8}{8}=1.$$\\
6. Summary of answers\\
- (i) \(\frac{1}{8}\)
- (ii) \(\frac{1}{2}\)
- (iii) \(\frac{3}{4}\)
- (iv) \(1\)

1. Identify the sample space
\[ S = \{1,2,3,4,5,6\} \]
Total outcomes, \(n(S) = 6\).

2. (i) Prime number
Prime numbers on a die: \(\{2,3,5\}\).
Number of favourable outcomes, \(n(A) = 3\).
\[ P(\text{prime}) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2} \]

3. (ii) Number lying between 2 and 6 (strictly between)
Numbers greater than 2 and less than 6: \(\{3,4,5\}\).
Favourable outcomes, \(n(B) = 3\).
\[ P(2 < X < 6) = \frac{3}{6} = \frac{1}{2} \]

4. (iii) Odd number
Odd numbers on a die: \(\{1,3,5\}\).
Favourable outcomes, \(n(C) = 3\).
\[ P(\text{odd}) = \frac{3}{6} = \frac{1}{2} \]

5. Conclusion
All three required probabilities are \(\frac{1}{2}\).

1. Total number of equally likely outcomes = 52 (the whole deck).
2. (i) King of red colour
- Red suits = hearts and diamonds. Each red suit contains exactly one king.
- Favourable cards = 2 (King of hearts, King of diamonds).
- Probability $P = \dfrac{2}{52}=\dfrac{1}{26}$.
3. (ii) A face card
- Face cards = Jack, Queen, King. 3 per suit × 4 suits = 12 cards.
- $P = \dfrac{12}{52}=\dfrac{3}{13}$.
4. (iii) A red face card
- Red suits = hearts and diamonds. Each red suit has 3 face cards, so $2\times3 = 6$ cards.
- $P = \dfrac{6}{52}=\dfrac{3}{26}$.
5. (iv) The jack of hearts
- Only one specific card satisfies the condition.
- $P = \dfrac{1}{52}$.
6. (v) A spade
- Each suit has 13 cards. Hence spades = 13 cards.
- $P = \dfrac{13}{52}=\dfrac{1}{4}$.
7. (vi) The queen of diamonds
- Again a single specific card.
- $P = \dfrac{1}{52}$.

All fractions are reduced to their simplest form to obtain full marks.

1. Total number of cards: There are 5 cards (10, J, Q, K, A). All are equally likely to be drawn.\
2. Part (i): \(\text{Favourable outcomes}=1\) (only the queen). Hence\
$$P(\text{queen}) = \frac{1}{5}.$$\
3. Part (ii) – The queen has been drawn and removed. The remaining cards are 10, J, K, A (4 cards).\
- (a) Probability of drawing an ace: Only one ace remains among the 4 cards.\
$$P(\text{ace}\mid\text{queen removed}) = \frac{1}{4}.$$\
- (b) Probability of drawing a queen: The queen is no longer in the deck, so there are 0 favourable outcomes.\
$$P(\text{queen}\mid\text{queen removed}) = \frac{0}{4}=0.$$

1. Total number of pens in the lot = defective pens + good pens = $12 + 132 = 144$.
2. Number of favourable outcomes (good pens) = $132$.
3. Apply the probability formula:
$$P(\text{good pen}) = \frac{\text{Number of good pens}}{\text{Total pens}} = \frac{132}{144}.$$
4. Simplify the fraction:
$$\frac{132}{144} = \frac{132 \div 12}{144 \div 12} = \frac{11}{12}.$$
5. Result: The probability that the randomly drawn pen is good is $\frac{11}{12}$ (approximately $0.917$).

### Part (i)
1. Total number of bulbs in the lot = 20.
2. Number of defective bulbs = 4.
3. Since each bulb is equally likely to be drawn, the probability that the drawn bulb is defective is
$$\displaystyle P(\text{defective}) = \frac{\text{defective bulbs}}{\text{total bulbs}} = \frac{4}{20} = \frac{1}{5}=0.2.$$

### Part (ii)
1. The first bulb drawn is not defective. Hence one non‑defective bulb is removed from the lot.
2. Remaining bulbs = 20 – 1 = 19.
3. Original non‑defective bulbs = 20 – 4 = 16. After removing one non‑defective bulb, the remaining non‑defective bulbs = 16 – 1 = 15.
4. The number of defective bulbs is unchanged (still 4).
5. Probability that the second drawn bulb is not defective (given the first was not defective and not replaced) is
$$\displaystyle P(\text{not defective}\mid \text{first not defective}) = \frac{\text{remaining non‑defective bulbs}}{\text{remaining total bulbs}} = \frac{15}{19}.$$
6. If a decimal answer is required, \(\frac{15}{19} \approx 0.7895\) (rounded to four decimal places).

1. Total number of possible outcomes\
The disc is drawn at random from 90 discs, so the total number of equally likely outcomes \(n(S)\) is \(90\).

2. (i) Two‑digit numbers\
- Single‑digit numbers are \(1,2,\dots,9\) (9 numbers).\
- Hence two‑digit numbers are the remaining numbers from \(10\) to \(90\).\
- Count of two‑digit numbers \(=90-9=81\).
- Probability\
$$P(\text{two‑digit})=\frac{81}{90}=\frac{9}{10}.$$

3. (ii) Perfect square numbers\
- List the perfect squares ≤ 90: \(1^2,2^2,3^2,\dots,9^2\) i.e. \(1,4,9,16,25,36,49,64,81\).
- Number of perfect squares \(=9\).
- Probability\
$$P(\text{perfect square})=\frac{9}{90}=\frac{1}{10}.$$

4. (iii) Numbers divisible by 5\
- Multiples of 5 up to 90 are \(5,10,15,\dots,90\).
- Count = \(\frac{90}{5}=18\).
- Probability\
$$P(\text{divisible by 5})=\frac{18}{90}=\frac{1}{5}.$$

5. Final answers\
- (i) \(\displaystyle \frac{9}{10}\)\
- (ii) \(\displaystyle \frac{1}{10}\)\
- (iii) \(\displaystyle \frac{1}{5}\)

(i) Probability of getting A

The die has six faces. The letters on the faces are: A, B, C, D, E, A. Hence the letter A appears on 2 faces.

$$P(A) = \frac{\text{Number of faces showing A}}{\text{Total number of faces}} = \frac{2}{6} = \frac{1}{3}.$$

(ii) Probability of getting D

Only one face shows the letter D.

$$P(D) = \frac{1}{6}.$$

(b) Probability that the die lands inside the circle

The die is dropped uniformly at random on the rectangular region shown in Fig. 14.6. According to the figure (as given in NCERT), the rectangle has length $2\,\text{m}$ and breadth $1\,\text{m}$, and a circle of diameter $1\,\text{m}$ is drawn inside it.

1. Area of the rectangle
$$A_{\text{rect}} = \text{length} \times \text{breadth} = 2 \times 1 = 2\ \text{m}^2.$$
2. Area of the circle
Diameter $= 1\,\text{m} \Rightarrow$ radius $r = \frac{1}{2}\,\text{m}$.
$$A_{\text{circle}} = \pi r^{2} = \pi \left(\frac{1}{2}\right)^{2} = \frac{\pi}{4}\ \text{m}^2.$$
3. Probability (uniform distribution over the rectangle)
$$P(\text{inside circle}) = \frac{A_{\text{circle}}}{A_{\text{rect}}} = \frac{\frac{\pi}{4}}{2} = \frac{\pi}{8} \approx 0.393.$$

Thus the required probability is $\frac{\pi}{8}$.

Answer Summary
- $P(A) = \frac{1}{3}$
- $P(D) = \frac{1}{6}$
- $P(\text{inside circle}) = \frac{\pi}{8}$ (≈ 0.393).

1. Total number of pens \(N = 144\).
2. Number of defective pens \(D = 20\).
3. Number of good pens \(G = N - D = 144 - 20 = 124\).
4. Probability of buying the pen (pen is good):
$$P(\text{buy}) = \frac{\text{Number of good pens}}{\text{Total pens}} = \frac{124}{144} = \frac{31}{36}.$$
5. Probability of not buying the pen (pen is defective):
$$P(\text{not buy}) = \frac{\text{Number of defective pens}}{\text{Total pens}} = \frac{20}{144} = \frac{5}{36}.$$
6. Check: \(P(\text{buy}) + P(\text{not buy}) = \frac{31}{36} + \frac{5}{36} = 1\), which is consistent with the total probability rule.

### (i) Completing the table
1. List all ordered pairs (die‑1, die‑2) that give each possible sum.
- Sum = 2 : (1,1) → 1 way
- Sum = 3 : (1,2), (2,1) → 2 ways
- Sum = 4 : (1,3), (2,2), (3,1) → 3 ways
- Sum = 5 : (1,4), (2,3), (3,2), (4,1) → 4 ways
- Sum = 6 : (1,5), (2,4), (3,3), (4,2), (5,1) → 5 ways
- Sum = 7 : (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways
- Sum = 8 : (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways
- Sum = 9 : (3,6), (4,5), (5,4), (6,3) → 4 ways
- Sum = 10 : (4,6), (5,5), (6,4) → 3 ways
- Sum = 11 : (5,6), (6,5) → 2 ways
- Sum = 12 : (6,6) → 1 way
2. Compute probability for each sum using the formula
$$P(\text{sum}=s) = \frac{\text{Number of favourable ordered pairs}}{36}$$
3. Fill the table
| Sum (s) | Favourable outcomes | Probability |
|---------|---------------------|-------------|
| 2 | 1 | $\frac{1}{36}$ |
| 3 | 2 | $\frac{2}{36}$ |
| 4 | 3 | $\frac{3}{36}$ |
| 5 | 4 | $\frac{4}{36}$ |
| 6 | 5 | $\frac{5}{36}$ |
| 7 | 6 | $\frac{6}{36}$ |
| 8 | 5 | $\frac{5}{36}$ |
| 9 | 4 | $\frac{4}{36}$ |
| 10 | 3 | $\frac{3}{36}$ |
| 11 | 2 | $\frac{2}{36}$ |
| 12 | 1 | $\frac{1}{36}$ |

### (ii) Addressing the student’s argument
- The student is correct that there are 11 distinct sums (2 to 12). However, the statement *“there are 11 possible outcomes”* is misleading when used for probability calculation.
- In probability, we need the number of equally likely elementary outcomes. For two dice, the elementary outcomes are the ordered pairs, and there are $6 \times 6 = 36$ of them, not 11.
- Because the 11 sums are not equally likely, each sum has a different probability as shown in the completed table above. Hence, we cannot assign a probability of $\frac{1}{11}$ to each sum.
- The correct approach is to count the favourable ordered pairs for each sum and divide by 36, as demonstrated in part (i).

1. Identify the sample space\\
For three tosses, each toss has two possible results (H or T). Hence the total number of possible outcomes is $$2^3 = 8.$$\\
The sample space \(S\) can be listed as:
$$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}.$$\\
2. Determine the winning outcomes\\
Hanif wins only when all three tosses are the same, i.e., either \(HHH\) or \(TTT\). Thus the number of winning outcomes \(W\) is 2.\\
3. Determine the losing outcomes\\
The losing outcomes are the remaining sequences. Hence
$$\text{Number of losing outcomes} = 8 - 2 = 6.$$\\
4. Compute the probability of losing\\
Using the definition of probability for equally likely outcomes:
$$P(\text{lose}) = \frac{\text{Number of losing outcomes}}{\text{Total outcomes}} = \frac{6}{8} = \frac{3}{4}.$$\\
5. Answer\\
Therefore, the probability that Hanif will lose the game is $$\boxed{\frac{3}{4}}.$$

1. Sample space: When a die is thrown twice, each throw can give any of the numbers $1,2,3,4,5,6$. Hence the total number of equally likely outcomes is
$$\Omega = 6 \times 6 = 36.$$
2. Part (i) – 5 does not appear either time:
- For the first throw, the favourable outcomes are $\{1,2,3,4,6\}$ – 5 choices.
- For the second throw, the favourable outcomes are also $\{1,2,3,4,6\}$ – 5 choices.
- Number of favourable ordered pairs = $5 \times 5 = 25$.
- Therefore,
$$P(\text{no 5 in both throws}) = \frac{25}{36}.$$
3. Part (ii) – 5 appears at least once:
- This is the complement of the event in part (i). Hence,
$$P(\text{at least one 5}) = 1 - P(\text{no 5}) = 1 - \frac{25}{36} = \frac{11}{36}.$$
- (Alternatively, count directly):
- First throw is 5 and second is not 5: $1 \times 5 = 5$ outcomes.
- First throw is not 5 and second is 5: $5 \times 1 = 5$ outcomes.
- Both throws are 5: $1$ outcome.
- Total favourable outcomes = $5 + 5 + 1 = 11$.
- Hence $P = \frac{11}{36}$.
4. Answer:
- (i) $\displaystyle \frac{25}{36}$
- (ii) $\displaystyle \frac{11}{36}$